In: Statistics and Probability
The given data set contains the amount of time(in minutes) students in an online course spent studying an assignment in a certain semester. According to the historical study, the mean time students would spend on this section nationwide is 95 minutes. While a researcher thinks that this semester student might be using more time. Use ? = 0.05 level of significance. Keep at least three decimal places for numerical answers.
(1) State the null hypothesis and alternative hypothesis for this researcher.
(2) Find the test statistic.
(3) Use critical value approach OR p-value approach to test the hypothesis. (If you use by-hand computation, show your steps; if you use Excel, provide the output)
(4) State the decision, then in the context of the research, state the conclusion.
Time (in minutes) |
100 |
65 |
62 |
62 |
54 |
101 |
124 |
202 |
212 |
78 |
101 |
114 |
57 |
69 |
67 |
60 |
113 |
75 |
53 |
38 |
82 |
132 |
96 |
113 |
86 |
129 |
94 |
78 |
38 |
134 |
44 |
90 |
40 |
169 |
176 |
137 |
122 |
249 |
49 |
151 |
127 |
129 |
215 |
137 |
55 |
346 |
Values ( X ) | ||
100 | 85.361 | |
65 | 1957.098 | |
62 | 2231.5326 | |
62 | 2231.5326 | |
54 | 3051.3582 | |
101 | 67.8828 | |
124 | 217.8842 | |
202 | 8604.5846 | |
212 | 10559.8026 | |
78 | 975.8814 | |
101 | 67.8828 | |
114 | 22.6662 | |
57 | 2728.9236 | |
69 | 1619.1852 | |
67 | 1784.1416 | |
60 | 2424.489 | |
113 | 14.1444 | |
75 | 1172.316 | |
53 | 3162.8364 | |
38 | 5075.0094 | |
82 | 741.9686 | |
132 | 518.0586 | |
96 | 175.2738 | |
113 | 14.1444 | |
86 | 540.0558 | |
129 | 390.4932 | |
94 | 232.2302 | |
78 | 975.8814 | |
38 | 5075.0094 | |
134 | 613.1022 | |
44 | 4256.1402 | |
90 | 370.143 | |
40 | 4794.053 | |
169 | 3571.3652 | |
176 | 4457.0178 | |
137 | 770.6676 | |
122 | 162.8406 | |
249 | 19533.1092 | |
49 | 3628.7492 | |
151 | 1743.9728 | |
127 | 315.4496 | |
129 | 390.4932 | |
215 | 11185.368 | |
137 | 770.6676 | |
55 | 2941.88 | |
346 | 56055.7238 | |
Total | 5025 | 172278.371 |
Mean
Standard deviation
To Test :-
H0 :-
H1 :-
Test Statistic :-
t = 1.5608
Test Criteria :-
Reject null hypothesis if
Result :- Fail to reject null hypothesis
Decision based on P value
P - value = P ( t > 1.5608 ) = 0.0628
Reject null hypothesis if P value <
level of significance
P - value = 0.0628 > 0.05 ,hence we fail to reject null
hypothesis
Conclusion :- Fail to reject null hypothesis
There is insufficient evidence to support the claim that this semester student might be using more time at ? = 0.05 level of significance.